1

### AIPMT 2011 Prelims

If n = 6, the correct sequence for filling of electrons will be
A
$ns \to \left( {n - 2} \right)f \to \left( {n - 1} \right)d \to np$
B
$ns \to \left( {n - 1} \right)d \to \left( {n - 2} \right)f \to np$
C
$ns \to \left( {n - 2} \right)f \to np \to \left( {n - 1} \right)d$
D
$ns \to np\left( {n - 1} \right)d \to \left( {n - 2} \right)f$
2

### AIPMT 2011 Prelims

The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e., $\lambda$1 and $\lambda$2 will be
A
$\lambda$1 = $\lambda$2
B
$\lambda$1 = 2$\lambda$2
C
$\lambda$1 = 4$\lambda$2
D
$\lambda$1 = ${1 \over 2}$ $\lambda$2

## Explanation

E1 = ${{hc} \over {{\lambda _1}}}$ and E2 = ${{hc} \over {{\lambda _2}}}$

${{{E_1}} \over {{E_2}}}$ = ${{hc} \over {{\lambda _1}}} \times {{{\lambda _2}} \over {hc}}$ = ${{{\lambda _2}} \over {{\lambda _1}}}$

$\Rightarrow$ $25 \over 50$ = ${{{\lambda _2}} \over {{\lambda _1}}}$

$\Rightarrow$ $1 \over 2$ = ${{{\lambda _2}} \over {{\lambda _1}}}$ $\Rightarrow$ $\lambda _1$ = 2$\lambda _2$
3

### AIPMT 2011 Prelims

The total number of atomic orbitals in fourth energy level of an atom is
A
8
B
16
C
32
D
4

## Explanation

The total number of atomic orbitals in any energy level is n2.

$\therefore$ 42 = 16
4

### AIPMT 2010 Mains

A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be

(h = 6.6 $\times$ 10$-$34 J s)
A
6.6 $\times$ 10$-$32 m
B
6.6 $\times$ 10$-$34 m
C
1.0 $\times$ 10$-$35 m
D
1.0 $\times$ 10$-$32 m

## Explanation

According to the de-Broglie equation, $\lambda$ = ${h \over {mv}}$

$\therefore$ $\lambda$ = ${{6.6 \times {{10}^{ - 34}}} \over {0.66 \times 100}} = 1 \times {10^{ - 35}}\,m$